Set 7 due today Set 8 due April 18 C-3 due April 18 Exam May 7

Answers set 7: (1) Shortest Route L-P60=MIN L-S125

L B V S M P

Delete L-P, ADD PATHS FROM P L-S125 P-M (new) 60+10=70 =MIN P-V (new) 60+55=115

Delete P-M, add path from m P-V115=MIN L-S125 M-B (new) 70+50=120

Delete P-V, add path from V M-B120=MIN L-S125

Delete both M-B,V-B L-S125 L-P 60 L-P-M 70 L-P-V115 L-P-M-B120

Answer to (2) PERT

1 4 3 A B CD E F 2

ES AND EF ACTIVITYESEF A05 B08 C07 DEF (C)=77+2=9 EMAX[(EF(B), EF(D)]=9 9+3=12=MAX= E(x) F77+4=11

ANSWER TO (a) E(x)=12

LF,LS ACTIVITYLFLS AE(x)=12 B C D E F

LF,LS ACTIVITYLFLS A B C D EE(x)= =9 FE(x)=1212-4=8

LF,LS ACTIVITYLFLS A B C DLS(E)=9 EE(x)=1212-3=9 FE(x)=1212-4=8

LF,LS ACTIVITYLFLS A B C DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

LF.LS ACTIVITYLFLS AE(x)=12 B CMIN[LS(D) LS(F)]= MIN(7,8)=7 DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

LF,LS ACTIVITYLFLS AE(x)=1212-5=7 BLS(E)=99-8=1 CMIN[LS(D) LS(F)]= MIN(7,8)=7 7-7=0 DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

LF,LS ACTIVITYLFLS AE(x)=1212-5=7 BLS(E)=99-8=1 CMIN[LS(D) LS(F)]= MIN(7,8)=7 7-7=0 DLS(E)=99-2=7 EE(x)=1212-3=9 FE(x)=1212-4=8

SLACK ACTIVITYLSESLS-ES A707 B101 C000 D770 E990 F871

ANSWER TO (b) CRITICAL PATH: C-D-E

NORMAL TABLE Kinderman Supplement, p 58 Row 1.0 Col.00

Z = 1.00 Z.00 …

Answer to (c) P(finish project before deadline)=.84

Excel Class demo NOT same as assignment should be in memo format Describe critical activities in sentence